QUESTION
2. In the given figure, we have a constant force 2900 lb.-f (=128900 N) acting over the beam at a distance of 15ft (=4.57 m) from one end and a variable force which keeps on decreasing from 50 plf (=715 N/m) to 20 plf(=291.8 N/m).
SOLUTION
Since, the variable force keeps on increasing from the left-end to the right-end, we can assume it as a linear force
At x=0, N so, we have c=291.8
At x=7.62 m, N so we have m=55.53
Now, calculating the aggregate force due to the variable force we have:
=
=7059 N
Hence, it has a total force of 135959 N acting in downward direction and reaction forces at each end equal to 67979.5 N in upward direction.
In the above diagram the resultant downwardforceFr , reaction forces Ra and Rb ,sheer forces V and V’, bending moments M and M’ are shown though a cut in the crosssection.
using the formula for the sheer stress and for the bending stress is , ahere the trms have their meanings, Shear stress=1.05and the bensding stresss=.
1. The axial stress in the cantilevered beam would be only due to the vertically down 25 kips (=245 N) force. Hence the axial stress=2.53 N.
Calculating the moments M1along negative z axis
M2= along positive Z axis
M3=(12*9.8*10*3.04)=3575.04 Nm along negative Z axis.
Hence, total moment=3872.96 along positive Z axis. Hence the bending moment
3. The figure shown in the problem shows a force F=50 kips (=490 N). An equal reaction force would act on the point of contact of the rod at the end of which the force is applied. Thus the shear stress===26923 .Now, the rotational torque due to the force=
LA24
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